Calculating Potential Difference in an Electric Field with Given Field Equation

AI Thread Summary
The electric field is defined by the equation Ex = 2.2x + 5x² kN/C, and the task is to calculate the potential difference between points at x = -2.0 m and x = 2.0 m. The integral used for the calculation is ∫(5x² + 2.2x) from -2 to 2. The initial calculation yielded 27V, but the correct answer is -27V due to the negative sign in the relationship between electric field and potential difference, as expressed by the formula ΔV = -∫E·ds. The oversight was in neglecting the negative sign in the integral, which is crucial for accurate results.
chopnhack
Messages
53
Reaction score
3

Homework Statement


An electric field is given by Ex = 2.2x + 5x2kNC-1. Find the potential difference between two points on the x-axis at x = -2.0 m and x = 2.0 m

Homework Equations


∫5x2+2.2x from -2 to 2

The Attempt at a Solution


I got the answer of 27V.

Solution said - negative 27V

Can someone please explain?
Thanks
 
Physics news on Phys.org
chopnhack said:
∫5x2+2.2x from -2 to 2
Remember there is a negative sign in front of this integral since
$$\mathbf{E}=-\nabla V$$.
 
  • Like
Likes chopnhack
##\Delta V = -\int_{-2}^2{\vec{E} \cdot d\vec{s}}##
You left out the negative sign.
 
  • Like
Likes chopnhack
NFuller said:
Remember there is a negative sign in front of this integral since
$$\mathbf{E}=-\nabla V$$.
yes, sadly I recall it now...

giphy.gif
 
  • Like
Likes NFuller
I was so happy to have found the proper way of handling it, I didn't recall the formula!
Thanks all.
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Calculating eletric potential using line integral of electric field
Replies
1
Views
2K
Potential difference between two points in an electric field
Replies
1
Views
2K
Earthing of two parallel conducting plates
Replies
11
Views
1K
Electric field in a rotating rod in a magnetic field
Replies
3
Views
2K
Why Is Work Positive When Done Against the Electric Field?
Replies
22
Views
3K
Understanding the energy of a dipole in a uniform electric field
Replies
2
Views
2K
Why is electric field at the center of a charged disk not zero?
Replies
4
Views
3K
Find the electric field between 2 finite discs
Replies
16
Views
1K
Ion moving through electric potential difference and magnetic field
Replies
8
Views
1K
Confused about direction of electric field
Replies
7
Views
3K

Hot Threads

Recent Insights

Back
Top